$\ce{XeF4}$ reacts with $\ce{SbF5}$ to form $[\ce{XeF_m}]^{n+}[\ce{SbF_y}]^{z-}$ $m + n + y + z =$ ______

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · p-Block Elements

XeF X_{4}

reacts with

SbF X_{5}

to form

[XeF X_{m}]^{n +} [SbF X_{y}]^{z -}

m + n + y + z =

______

Answer: 11

Solution

Sb F_{5}

is a powerful fluoride acceptor, so it abstracts one

F^{-}

from

Xe F_{4}

. Losing a fluoride leaves the cation

[Xe F_{3}]^{+}

, and the departed fluoride adds to antimony to build the octahedral anion

[Sb F_{6}]^{-}

. Comparing with

[Xe F_{m}]^{n +} [Sb F_{y}]^{z -}

gives

m = 3

n = 1

y = 6

and

z = 1

. Their sum is

3 + 1 + 6 + 1 = 11

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Integer answer

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